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goldfiish [28.3K]
3 years ago
6

Which substance is most easily grounded?

Physics
2 answers:
swat323 years ago
5 0

Answer:

The reference point in an electrical circuit to measure voltages is called <em>Earth</em> or <em>Ground</em>; that point is a direct hookup to earth that helps stabilizing electrical currents to prevent high voltage surges.

What makes a substance easily grounded is to be a good energy conductor; metallic objects are generally good energy conductors aka GEM <em>"Ground Enhancement Materials"</em> and they are mostly produced from rocks and minerals. That's why the answer is (B.) <em>Iron Nail</em>.

Explanation:

Basile [38]3 years ago
3 0

Answer:Iron nail

Explanation:

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Explain the evidence that Neils Bohr used to confirm that the electrons in an atom orbit in fixed shells. (4)
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What must be the distance (in meters) between a point charge q1 = 16 μC and a point charge q2 = 32 μC for the electrostatic forc
adoni [48]

Answer:

d = 0.71 meters

Explanation:

It is given that,

Charge 1, q_1=16\ \mu C=16\times 10^{-6}\ C

Charge 2, q_2=32\ \mu C=32\times 10^{-6}\ C

Electrostatic force between charges, F = 9 N

Let d is the distance between the charges. The electrostatic force between the charges is given by the product of charges and divided by square of distance between them. Mathematically, it is given by :

F=k\dfrac{q_1q_2}{d^2}

d=\sqrt{\dfrac{kq_1q_2}{F}}

d=\sqrt{\dfrac{9\times 10^9\times 16\times 10^{-6}\times 32\times 10^{-6}}{9}}

d = 0.71 meters

So, the distance between the charges is 0.71 meters. Hence, this is the required solution.

6 0
4 years ago
A car accelerates uniformly from rest at a speed of 1.67 ft s^2 over a distance of 5 yards.What is the acceleration of a car?
Nina [5.8K]

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

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t= 30/23

t= 1.30 seconds

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5 0
3 years ago
Read 2 more answers
A rectangular plate has a length of (21.7 ± 0.2) cm and a width of (8.2 ± 0.1) cm. Calculate the area of the plate, including it
Grace [21]

Answer:

(177.94 ± 3.81) cm^2

Explanation:

l + Δl = 21.7 ± 0.2 cm

b + Δb = 8.2 ± 0.1 cm

Area, A = l x b = 21.7 x 8.2 = 177.94 cm^2

Now use error propagation

\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b}

\frac{\Delta A}{A}=\frac{0.2}{21.7}+\frac{0.1}{8.2}

\Delta A=177.94 \times \left ( 0.0092 + 0.0122 \right )=3.81

So, the area with the error limits is written as

A + ΔA = (177.94 ± 3.81) cm^2

8 0
3 years ago
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